It is given that f(x) = \(\dfrac{ (4x + 3)} {(6x – 4)}\), x ≠ \( \dfrac{2}{3}\)

Now we have to show fof(x) = x

(fof)(x) = f (f(x))

=\( \dfrac{ (4x+ 3)}{(6x – 4)}\)

= \( \dfrac{34x}{34}\)

= x

Therefore, fof(x) = x for all x ≠ \( \dfrac{2}{3}\)

= fof = 1

Hence, the given function f is invertible and the inverse of f is f itself

Answered by Aaryan | 1 year agoLet A be the set of first five natural and let R be a relation on A defined as follows: (x, y) R x ≤ y

Express R and R^{-1} as sets of ordered pairs. Determine also

**(i) **the domain of R^{‑1}

**(ii)** The Range of R.

A function f: R → R is defined as f(x) = x^{3} + 4. Is it a bijection or not? In case it is a bijection, find f^{−1} (3).

If f: R → R be defined by f(x) = x^{3} −3, then prove that f^{−1} exists and find a formula for f^{−1}. Hence, find f^{−1 }(24) and f^{−1} (5).

Consider f: R^{+} → [−5, ∞) given by f(x) = 9x^{2} + 6x − 5. Show that f is invertible with f^{-1}(x) = \(
\dfrac{\sqrt{(x+6)-1}}{3}\)

Consider f: R → R^{+} → [4, ∞) given by f(x) = x^{2} + 4. Show that f is invertible with inverse f^{−1} of f given by f^{−1}(x) = \( \sqrt{ (x-4)}\) where R^{+} is the set of all non-negative real numbers.